NCERT Solutions For Class 6 Maths Chapter 10 - Mensuration
NCERT Solutions for Class 6 Maths offer comprehensive explanations for the questions found within the NCERT textbooks endorsed by the Central Board of Secondary Education (CBSE). Orchids the international school provides these NCERT Class 6 Maths Solutions on a chapter-by-chapter basis, aiming to assist students in resolving any uncertainties and acquiring a profound comprehension of the subject matter. These resources, including NCERT Solutions, are conveniently accessible in PDF format, allowing students to download them for offline learning.
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The NCERT Solutions For Class 6 Maths Chapter 10 - Mensuration are tailored to help the students master the concepts that are key to success in their classrooms. The solutions given in the PDF are developed by experts and correlate with the CBSE syllabus of 2023-2024. These solutions provide thorough explanations with a step-by-step approach to solving problems. Students can easily get a hold of the subject and learn the basics with a deeper understanding. Additionally, they can practice better, be confident, and perform well in their examinations with the support of this PDF.
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Access Answers to NCERT Solutions For Class 6 Maths Chapter 10 - Mensuration
Students can access the NCERT Solutions For Class 6 Maths Chapter 10 - Mensuration. Curated by experts according to the CBSE syllabus for 2023–2024, these step-by-step solutions make Maths much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.
Introduction
What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm, respectively?
The required length of wooden strip = Perimeter of photograph
= 2 (Length + Breadth)
= 2 (32 + 21)
= 2 (53)
= 2 × 53
= 106 cm
∴ The required length of the wooden strip is 106 cm
A piece of string is 30 cm long. What will be the length of each side if the string is used to form:
(a) a square?
(b) an equilateral triangle?
(c) a regular hexagon?
(a) Perimeter of square = 30 cm
4 × side = 30
Side = 30 / 4
Side = 7.5 cm
(b) Perimeter of an equilateral triangle = 30 cm
3 × side = 30
Side = 30 / 3
Side = 10 cm
(c) Perimeter of a regular hexagon = 30 cm
6 × side = 30
Side = 30 / 6
Side = 5 cm
A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?
Perimeter of the field = 2 (Length + Breadth)
= 2 (0.7 + 0.5)
= 2 (1.2)
= 2 × 1.2
= 2.4 km
Each side is to be fenced with 4 rows = 4 × 2.4
= 9.6 km
∴ The total length of the required wire is 9.6 km
Find the perimeter of each of the following figures:
(a) Perimeter = Sum of all the sides
= 1 + 2 + 4 + 5
= 12 cm
(b) Perimeter = Sum of all the sides
= 23 + 35 + 35 + 40
= 133 cm
(c) Perimeter = Sum of all the sides
= 15 + 15 + 15 + 15
= 60 cm
(d) Perimeter = Sum of all the sides
= 4 + 4 + 4 + 4 + 4
=20 cm
(e) Perimeter = Sum of all the sides
= 1 + 4 + 0.5 + 2.5 + 2.5 + 0.5 + 4
= 15 cm
(f) Perimeter = Sum of all the sides
= 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3
= 52 cm
The lid of a rectangular box of sides 40 cm by 10 cm is sealed all around with tape. What is the length of the tape required?
Length of required tape = Perimeter of rectangle
= 2 (Length + Breadth)
= 2 (40 + 10)
= 2 (50)
= 100 cm
∴ The required length of tape is 100 cm
A table top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table top?
Length of table top = 2 m 25 cm = 2.25 m
Breadth of table top = 1 m 50 cm = 1.50 m
Perimeter of table top = 2 (Length + Breadth)
= 2 (2.25 + 1.50)
= 2 (3.75)
= 2 × 3.75
= 7.5 m
∴ The perimeter of the table top is 7.5 m
Find the perimeter of each of the following shapes:
(a) A triangle of sides 3 cm, 4 cm and 5 cm
(b) An equilateral triangle of side 9 cm
(c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.
(a) Perimeter of triangle = 3 + 4 + 5
= 12 cm
(b) Perimeter of an equilateral triangle = 3 × side
= 3 × 9
= 27 cm
(c) Perimeter of isosceles triangle = 8 + 8 + 6
= 22 cm
Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
Perimeter of triangle = 10 + 14 + 15
= 39 cm
∴ The perimeter of the triangle is 39 cm
Find the perimeter of a regular hexagon with each side measuring 8 m.
Perimeter of hexagon = 6 × 8
= 48 m
∴ The perimeter of the regular hexagon is 48 m
Find the side of the square whose perimeter is 20 m.
Perimeter of square = 4 × side
20 = 4 × side
Side = 20 / 4
Side = 5 m
∴ The side of the square is 5 m
The perimeter of a regular pentagon is 100 cm. How long is its each side?
The perimeter of the regular pentagon = 100 cm
5 × side = 100 cm
Side = 100 / 5
Side = 20 cm
∴ The side of the pentagon is 20 cm
Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?
Let x cm be the third side
Perimeter of triangle = 36 cm
12 + 14 + x = 36
26 + x = 36
x = 36 – 26
x = 10 cm
∴ The third side is 10 cm
Find the cost of fencing a square park of side 250 m at the rate of ₹ 20 per metre.
Side of square = 250 m
Perimeter of square = 4 × side
= 4 × 250
= 1000 m
Cost of fencing = ₹ 20 per m
Cost of fencing for 1000 m = ₹ 20 × 1000
= ₹ 20,000
Find the cost of fencing a rectangular park of length 175 cm and breadth 125 m at the rate of ₹ 12 per metre.
Length = 175 cm
Breadth = 125 m
Perimeter of rectangular park = 2 (Length + Breadth)
= 2 (175 + 125)
= 2 (300)
= 2 × 300
= 600 m
Cost of fencing = 12 × 600
= 7200
∴ The cost of fencing is ₹ 7,200
Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with length 60 m and breadth 45 m. Who covers less distance?
Perimeter of square = 4 × side
= 4 × 75
= 300 m
∴ The distance covered by Sweety is 300 m
Perimeter of rectangular park = 2 (Length + Breadth)
= 2 (60 + 45)
= 2 (105)
= 2 × 105
= 210 m
∴ The distance covered by Bulbul is 210 m
Hence, Bulbul covers less distance than Sweety.
What is the perimeter of each of the each of the following figures? What do you infer from the the answers?
(a) Perimeter of square = 4 × side
= 4 × 25
= 100 cm
(b) Perimeter of rectangle = 2 (40 + 10)
= 2 × 50
= 100 cm
(c) Perimeter of rectangle = 2 (Length + Breadth)
= 2 (30 + 20)
= 2 (50)
= 2 × 50
= 100 cm
(d) Perimeter of triangle = 30 + 30 + 40
= 100 cm
All the figures have the same perimeter.
Avneet buys 9 square paving slabs, each with a side of 1 / 2 m. He lays them in the form of a square.
(a) What is the perimeter of his arrangement [fig 10.7(i)]?
(b) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement [(Fig 10.7 (ii)]?
(c) Which has greater perimeter?
(d) Avneet wonders if there is a way of getting an even greater perimeter. Can you find a way of doing this? (The paving slabs must meet along complete edges i.e they cannot be broken.)
(a) Side of square = 3 × side
= 3 × 1 / 2
= 3 / 2 m
Perimeter of Square = 4 × 3 / 2
= 2 × 3
= 6 m
(b) Perimeter = 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1
= 10 m
(c) The arrangement in the form of a cross has greater perimeter
(d) Perimeters greater than 10 m cannot be determined.
Perimeter
Find the areas of the following figures by counting squares.
(a) The figure contains only 9 fully-filled squares. Hence, the area of this figure will be 9 square units.
(b) The figure contains only 5 fully-filled squares. Hence, the area of this figure will be 5 square units.
(c) The figure contains 2 fully-filled squares and 4 half-filled squares. Hence, the area of this figure will be 4 square units.
(d) The figure contains only 8 fully-filled squares. Hence, the area of this figure will be 8 square units.
(e) The figure contains only 10 fully-filled squares. Hence, the area of this figure will be 10 square units.
(f) The figure contains only 2 fully-filled squares and 4 half-filled squares. Hence, the area of this figure will be 4 square units.
(g) The figure contains 4 fully-filled squares and 4 half-filled squares. Hence, the area of this figure will be 6 square units.
(h) The figure contains 5 fully-filled squares. Hence, the area of this figure will be 5 square units.
(i) The figure contains 9 fully-filled squares. Hence, the area of this figure will be 9 square units.
(j) The figure contains 2 fully-filled squares and 4 half-filled squares. Hence, the area of this figure will be 4 square units.
(k) The figure contains 4 fully-filled squares and 2 half-filled squares. Hence, the area of this figure will be 5 square units.
(l) From the given figure, we observe
|
Covered area |
Number |
Area estimate (square units) |
|
Fully-filled squares |
2 |
2 |
|
Half-filled squares |
– |
– |
|
More than half-filled squares |
6 |
6 |
|
Less than half-filled squares |
6 |
0 |
Therefore, the total area = 2 + 6
= 8 square units
(m) From the given figure, we observe
|
Covered area |
Number |
Area estimate (square units) |
|
Fully-filled squares |
5 |
5 |
|
Half-filled squares |
– |
– |
|
More than half-filled squares |
9 |
9 |
|
Less than half-filled squares |
12 |
0 |
Therefore, the total area = 5 + 9
= 14 square units
(n) From the given figure, we observe
|
Covered area |
Number |
Area estimate (square units) |
|
Fully-filled squares |
8 |
8 |
|
Half-filled squares |
– |
– |
|
More than half-filled squares |
10 |
10 |
|
Less than half-filled squares |
9 |
0 |
Therefore, the total area = 8 + 10 = 18 square units
Area
Find the areas of the squares whose sides are:
(a) 10 cm
(b) 14 cm
(c) 5 m
(a) Area of square = side2
= 102
= 100 cm2
(b) Area of square = side2
= 142
= 196 cm2
(c) Area of square = side2
= 52
=25 cm2
The length and breadth of the three rectangles are as given below:
(a) 9 m and 6 m
(b) 17 m and 3 m
(c) 4 m and 14 m
Which one has the largest area, and which one has the smallest?
(a) Area of rectangle = l × b
= 9 × 6
= 54 m2
(b) Area of rectangle = l × b
= 17 × 3
= 51 m2
(c) Area of rectangle = l × b
= 4 × 14
= 56 m2
Option (c), the rectangle with an area of 56 m2, has the largest area, and option (b), the rectangle with an area of 51 m2, has the smallest area.
The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden.
Area of rectangle = length × width
300 = 50 × width
width = 300 / 50
width = 6 m
∴ The width of the garden is 6 m.
What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of ₹ 8 per 100 sq m?
Area of land = length × breadth
= 500 × 200
= 1,00,000 m2
∴ Cost of tiling 1,00,000 sq m of land = (8 × 1,00,000) / 100
= ₹ 8000
A table top measures 2 m by 1 m 50 cm. What is its area in square metres?
Given
l = 2 m
b = 1 m 50 cm = 1.50 m
Area = l × b = 2 × 1.50
= 3 m2
A room is 4 m long and 3 m 50 cm wide. How many square metres of carpet are needed to cover the floor of the room?
Given
l = 4m
b = 3 m 50 cm = 3.50 m
Area = l × b = 4 × 3.50
= 14 m2
A floor is 5 m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find the area of the floor that is not carpeted.
Area of floor = l × b = 5 × 4
= 20 m2
Area of square carpet = 3 × 3
= 9 m2
Area of floor that is not carpeted = 20 – 9
= 11 m2
∴ The area of the floor that is not carpeted is 11 m2.
Five square flower beds, each of sides 1 m, are dug on a piece of land 5 m long and 4 m wide. What is the area of the remaining part of the land?
Area of flower square bed = 1 × 1
= 1 m2
Area of 5 square bed = 1 × 5
= 5 m2
Area of land = 5 × 4
= 20 m2
Remaining part of the land = Area of land – Area of 5 square bed
= 20 – 5
= 15 m2
∴ The remaining part of the land is 15 m2.
By splitting the following figures into rectangles, find their areas (The measures are given in centimetres).
(a)
Area of yellow region = 3 × 3
= 9 cm2
Area of orange region = 1 × 2
= 2 cm2
Area of grey region = 3 × 3
= 9 cm2
Area of brown region = 2 × 4
= 8 cm2
Total area = 9 + 2 + 9 + 8
= 28 cm2
∴ The total area is 28 cm2.
(b)
Area of brown region = 3 × 1
= 3 cm2
Area of orange region = 3 × 1
= 3 cm2
Area of grey region = 3 × 1
= 3 cm2
Total area = 3 + 3 + 3
= 9 cm2
∴ The total area is 9 cm2.
Split the following shapes into rectangles and find their areas. (The measures are given in centimetres)
(a)
Total area of the figure = 12 × 2 + 8 × 2
= 40 cm2
(b)
There are 5 squares, and each side is 7 cm.
Area of 5 squares = 5 × 72
= 245 cm2
(c)
Area of grey rectangle = 2 × 1
= 2 cm2
Area of brown rectangle = 2 × 1
= 2 cm2
Area of orange rectangle = 5 × 1
= 5 cm2
Total area = 2 + 2 + 5
= 9 cm2
How many tiles whose length and breadth are 12 cm and 5 cm, respectively, will be needed to fit in a rectangular region whose length and breadth are respectively,
(a) 100 cm and 144 cm?
(b) 70 cm and 36 cm?
(a) Area of rectangle = 100 × 144
= 14400 cm
Area of one tile = 5 × 12
= 60 cm2
Number of tiles = (Area of rectangle) / (Area of one tile)
= 14400 / 60
= 240
Hence, 240 tiles are needed.
(b) Area of rectangle = 70 × 36
= 2520 cm2
Area of one tile = 5 × 12
= 60 cm2
Number of tiles = (Area of rectangle) / (Area of one tile)
= 2520 / 60
= 42
Hence, 42 tiles are needed.
Find the area of the rectangles whose sides are:
(a) 3 cm and 4 cm
(b) 12 m and 21 m
(c) 2 km and 3 km
(d) 2 m and 70 cm
We know that
Area of rectangle = length × breadth
(a) l = 3 cm and b = 4 cm
Area = l × b = 3 × 4
= 12 cm2
(b) l = 12 m and b = 21 m
Area = l × b = 12 × 21
= 252 m2
(c) l = 2 km and b = 3 km
Area = l × b = 2 × 3
= 6 km2
(d) l = 2 m and b = 70 cm = 0.70 m
Area = l × b = 2 × 0.70
= 1.40 m2
Frequently Asked Questions
The NCERT solution for Class 6 Chapter 10: Mensuration is important as it provides a structured approach to learning, ensuring that students develop a strong understanding of foundational concepts early in their academic journey. By mastering these basics, students can build confidence and readiness for tackling more difficult concepts in their further education.
Yes, the NCERT solution for Class 6 Chapter 10: Mensuration is quite useful for students in preparing for their exams. The solutions are simple, clear, and concise allowing students to understand them better. They can solve the practice questions and exercises that allow them to get exam-ready in no time.
You can get all the NCERT solutions for Class 6 Maths Chapter 10 from the official website of the Orchids International School. These solutions are tailored by subject matter experts and are very easy to understand.
Yes, students must practice all the questions provided in the NCERT solution for Class 6 Maths Chapter 10: Mensuration as it will help them gain a comprehensive understanding of the concept, identify their weak areas, and strengthen their preparation.
Students can utilize the NCERT solution for Class 6 Maths Chapter 10 effectively by practicing the solutions regularly. Solve the exercises and practice questions given in the solution.